> template<typename K, typename V> static QVector<K>
> keysToVector(QHash<K,V> const& hash)
> {
> QVector<K> v;
> v.reserve(hash.size());
>
> for (auto it = hash.keyBegin(); it != hash.keyEnd(); ++it)
> v.append(*it);
>
> return v;
> }
>
> QVector<int> CompatibilityInterfaceImpl::getActionIds() const
> {
> return keysToVector(_actions);
> }
Dear all,
I am sorry to kind of hijack this thread with a follow-up question, but I am
trying to better understand what‘s going on „under the C++ hood“, especially
given the solution above.
When we look at the original code again:
QVector<int> CompatibilityInterfaceImpl::getActionIds() const
{
return _actions.keys().toVector(); // _actions is a QHash
}
And the warning it generates:
allocating an unneeded temporary container [clazy-container-anti-pattern]
I am having trouble to see the difference of those two solutions, in the
context of the warning.
I do understand that in the original solution a vector is allocated within the
toVector() method, which is then returned (by value) and hence copied into a
„temporary variable“ (it is not assigned to a declared variable, that is my
understanding of „temporary“ - correct?). That „temporary variable“ shortly
lives in the scope of getActionIds().
As its value needs to be returned by getActionIds() it (the vector) is copied
once more, into the final result variable (of the caller of getActionIds().
So I (naively?) count:
* One vector allocation
* Two vector copies by value
Now in the second solution I count (wrongly?) the same operations:
* One vector allocation (in keysToVector)
* Two vector copies (return by value)
So why then is „return keysToVector()“ more efficient than the first „return
map.keys().toVector()“? Is this related to the C++11 „move“ operator where the
compiler can „optimise away“ the „copy“ operation in the „return by value“? If
that is the case, why can‘t the same optimisation be done in the original case?
Or did I simply miscount the number of allocate/copy operations?
Thank you!
Oliver
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